题目

Problem

A random sample of 50 values of the continuous random variable $X$ was taken. These values are summarised in the following table.

Interval	$1 \leq x < 1.5$	$1.5 \leq x < 2$	$2 \leq x < 2.5$	$2.5 \leq x < 3$	$3 \leq x < 3.5$	$3.5 \leq x \leq 4$
Observed frequency	3	3	8	11	13	12

It is required to test the goodness of fit of the distribution with probability density function f given by

f(x) = \begin{cases} \dfrac{1}{24}\bigg(\dfrac{4}{x^2} + x^2\bigg), & 1 \leq x \leq 4,\\ 0, & \text{otherwise.} \end{cases}

The expected frequencies, correct to 4 decimal places, are given in the following table.

Interval	$1 \leq x < 1.5$	$1.5 \leq x < 2$	$2 \leq x < 2.5$	$2.5 \leq x < 3$	$3 \leq x < 3.5$	$3.5 \leq x \leq 4$
Expected frequency	4.4271	$a$	6.1285	8.4549	$b$	14.9678

(a) Show that $a = 4.6007$ and find the value of $b$ .

[3]

(b) Carry out a goodness of fit test, at the 10% significance level, to test whether f is a satisfactory model for the data.

[6]

题目中文翻译

从连续随机变量 $X$ 中随机抽取 50 个值。这些值汇总在下表中。

区间	$1 \leq x < 1.5$	$1.5 \leq x < 2$	$2 \leq x < 2.5$	$2.5 \leq x < 3$	$3 \leq x < 3.5$	$3.5 \leq x \leq 4$
观察频数	3	3	8	11	13	12

现在要检验具有以下概率密度函数 f 的分布的拟合优度：

f(x) = \begin{cases} \dfrac{1}{24}\bigg(\dfrac{4}{x^2} + x^2\bigg), & 1 \leq x \leq 4,\\ 0, & \text{otherwise.} \end{cases}

下表给出了期望频数，结果精确到 4 位小数。

区间	$1 \leq x < 1.5$	$1.5 \leq x < 2$	$2 \leq x < 2.5$	$2.5 \leq x < 3$	$3 \leq x < 3.5$	$3.5 \leq x \leq 4$
期望频数	4.4271	$a$	6.1285	8.4549	$b$	14.9678

(a) 证明 $a = 4.6007$ ，并求 $b$ 的值。

(3)

(b) 在 10% 显著性水平下进行拟合优度检验，检验 f 是否是该数据的一个令人满意的模型。

(6)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

期望频数等于区间概率乘样本量。对密度求原函数后，在两个缺失区间分别代入端点。

答题过程

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For an interval $[u,v]$ inside the support, the expected frequency is

\frac{50}{24}\Big[-\frac4x+\frac{x^3}{3}\Big]_u^v.

Substituting the endpoints gives

\begin{align*} a=&\,\frac{50}{24}\Big[\Big(-2+\frac83\Big)\\ &\,\hspace{2pt}-\Big(-\frac83+\frac98\Big)\Big]\\ =&\,4.600694\ldots\approx4.6007,\\ b=&\,\frac{50}{24}\Big[\Big(-\frac87+\frac{343}{24}\Big)\\ &\,\hspace{2pt}-\Big(-\frac43+9\Big)\Big]\\ =&\,11.421130\ldots\approx11.4211. \end{align*}

Both answers are given to four decimal places.

解法二

思路

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先积分算出题目要求证明的 $a$ ，再利用所有期望频数之和为样本量求 $b$ 。

答题过程

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For the second interval,

\begin{align*} a=&\,\frac{50}{24}\int_{1.5}^2\Big(\frac4{x^2}+x^2\Big)\,\mathrm{d}x\\ =&\,\frac{1325}{288}=4.600694\ldots\approx4.6007. \end{align*}

The expected frequencies sum to $50$ . Using their displayed values,

\begin{align*} b\approx&\,50-(4.4271+4.6007+6.1285\\ &\,\hspace{2pt}+8.4549+14.9678)\\ =&\,11.4210. \end{align*}

The slight difference in the last decimal comes from the rounded frequencies; integrating the density gives $b=11.4211$ to four decimal places.

(b)

解法一

思路

展开

前两个格子的期望频数都低于 $5$ ，先合并。密度参数已给定，按合并后的组数计算自由度。

答题过程

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Test $H_0$ that the stated density fits the observations against $H_1$ that it does not. Combining the first two intervals gives observed count $6$ and expected count $9.02777\ldots$ . The remaining counts have expected values above $5$ . The statistic from the five groups is

\begin{align*} &\,\chi^2=\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(6-9.02777\ldots)^2}{9.02777\ldots}\\ +&\,\frac{(8-6.12847\ldots)^2}{6.12847\ldots}+\cdots\\ =&\,3.15985\ldots. \end{align*}

There are $5-1=4$ degrees of freedom. The upper $10\%$ critical value is $7.779$ . Since the statistic is smaller, we fail to reject $H_0$ . There is insufficient evidence at the $10\%$ level that the stated density fails to fit the observations.

CIE 9231 2023 June Paper 42 Q3

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程